solutions to chapter 11 exercise problems problem...

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Solutions to Chapter 11 Exercise Problems

Problem 11.1

Two helical gears are cut with a spur gear hob that has a diametral pitch of 4 and a pressure angle of 20˚. The pinion has 15 teeth, the gear has 35 teeth, and the helix angle is 30˚. Determine the minimum recommended face width. Using the minimum face width, find the transverse diametral pitch, the pitch cylinder radii, and the axial, transverse, and total contact ratios.

Solution

The limiting condition for the face width is given by Eq. (11.15)

F1.15pnsin

The normal pitch is given by

pn = Pn

Therefore,

F 1.15Pnsin

Now1.15Pn sin

= 1.154sin30

=1.8064

Therefore,

F 1.8064

The properties of the gears in the normal direction will be the same as those of the hob. The transverse diametral pitch (Pt) for both gears is related to the normal diametral pitch by Eq. (11.5). Based on the diametral pitch of the hob,

Pt = Pncos = 4cos30 = 3.464

The pitch cylinder diameters are related to the diametral pitch through Eq. (11.4) Therefore,

Dt =NtPt

and the pitch cylinder radius for the pinion is

r2t =N2t2Pt

= 152(3.464)

= 2.165 in

The pitch cylinder radius for the gear is

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r3t =N3t2Pt

= 352(3.464)

= 5.052 in

The transverse contact ratio is given by Eq. (11.16) as

mct =pt( 2 + 3)cos t

where i = rpi sin t + ai

2 +2airpi + rpi2 sin2 t , i = 2, 3

The addenda are determined by the hob. Because a standard hob is used, both addenda are given by

a = 1Pn= 14= 0.25 inch

The transverse circular pitch is given by Eq. (11.3) as

pt = Pt=3.464

= 0.907

and the transverse pressure angle is given by Eq. (11.7) as

t = tan 1 tan n

cos

= tan 1 tan20

cos30( ) = 22.796̊

Now,2 = rp2sin t + a2

2 + 2a2rp2 + rp22 sin2 t

= 2.165sin(22.796)+ (0.25)2 + 2(0.25)(2.165) + [2.165sin(22.796)]2 = 0.5208and

3 = rp3sin t + a32 +2a3rp3 + rp3

2 sin2 t

= 5.0518sin(22.796)+ (0.25)2 + 2(0.25)(5.0518) + [5.0518sin(22.796)]2 = 0.5764

Therefore, the transverse contact ratio is


=0.907(0.5208 +0.5764)

cos(22.796)=1.3123

The axial contact ratio is given by Eq. (11.16) as

mca = F tanpt

=1.8064tan30

0.907=1.1500

The total contact ratio is

mc = mct +mca = 2.4623+1.1500 = 2.4623

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Problem 11.2

Two helical gears are cut with the same tooth numbers and with the same cutter as given in Problem 11.1. The helix angle is 30˚. Find the transverse pressure angle, the transverse diametral pitch, and the axial pitch.

Solution

The transverse pressure angle is given by Eq. (11.7) as

tan t = tan 1 tan n

cos

= tan 1 tan20˚

cos30̊( ) = tan 1 0.4203( ) = 22.79̊

The transverse diametral pitch (Pt) for both gears is related to the normal diametral pitch by Eq. (11.5). Based on the diametral pitch of the hob,

Pt = Pncos = 4cos30 = 3.464

The axial pitch is given by Eq. (11.6). Then

pa =pnsin

=Pn sin

=4sin30̊

=1.5708

Problem 11.3

Two parallel helical gears are cut with a 20˚ normal pressure angle and a 45˚ helix angle. They have a diametral pitch of 12 in the normal plane and have 10 and 41 teeth, respectively. Find the transverse pressure angle, transverse circular pitch, and transverse diametral pitch. Also determine the minimum face width, and using that face width, determine the total contact ratio.

Solution


t = tan 1 tan n

cos

= tan 1 tan20̊

cos45̊( ) = tan 1 0.5147( ) = 27.24̊

The transverse diametral pitch (Pt) for both gears is related to the normal diametral pitch by Eq. (11.5). Based on the diametral pitch of the hob,

Pt = Pncos =12cos45 = 8.485


pt = Pt=8.485

= 0.3702


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F1.15pnsin


pn = Pn

Therefore,

F 1.15Pnsin

Now

1.15Pn sin

= 1.1512sin 45

= 0.4258

Therefore,

F 0.4258

The pitch cylinder diameters are related to the diametral pitch through Eq. (11.4) Therefore,

Dt =NtPt

and the pitch cylinder radius for the pinion is

r2t =N2t2Pt

= 102(8.485)

= 0.589 in

The pitch cylinder radius for the gear is

r3t =N3t2Pt

= 412(8.485)

= 2.416 in

The transverse contact ratio is given by Eq. (11.16) as


where i = rpi sin t + ai

2 +2airpi + rpi2 sin2 t , i = 2, 3

The addenda are determined by the hob. Because a standard hob is used, both addenda are given by

a = 1Pn= 112

= 0.08333 in

Now,

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2 = r2t sin t + a22 +2a2r2t + r2t2 sin2 t

= 0.589sin(27.24) + (0.08333)2 + 2(0.08333)(0.589) + [0.589sin(27.24)]2 = 0.1521and

3 = r3t sin t + a32 +2a3r3t + r3t2 sin2 t

= 2.416sin(27.24) + (0.08333)2 + 2(0.08333)(2.416) + [2.416sin(27.24)]2 = 0.1719

Therefore, the transverse contact ratio is


=0.3702(0.1521+ 0.1719)

cos(27.24)= 0.9841

The axial contact ratio is given by Eq. (11.16) as

mca = F tanpt

=0.4258tan45

0.3702= 1.1500

The total contact ratio is

mc = mct +mca = 0.9841+1.1500 = 2.1341

Problem 11.4

A helical gear has 18 teeth and a transverse diametral pitch of 6. The face width is 1.5, and the helix angle is 25˚. Determine the axial pitch, normal pitch, lead, transverse pitch diameter, and minimum face width.

Solution


t = tan 1 tan n

cos

= tan 1 tan20˚

cos25̊( ) = tan 1 0.9063( ) = 21.88̊

The transverse diametral pitch (Pt) for the gear is related to the normal diametral pitch by Eq. (11.5). Based on the transverse diametral pitch

Pn = Pt / cos = 6 / cos25 = 6.620

The transverse pitch diameter is given by

dt =N2Pt= 186= 3


pt = Pt=6= 0.5236

The axial pitch is given by Eq. (11.6). Then

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pa =pnsin

=Pn sin

=6.620sin25̊

=1.1229

The lead is given by Eq. (11.25) as

L = N2pa =18(1.1229) = 20.212 in


F1.15pnsin


pn = Pn

Therefore,

Fmin 1.15Pn sin

Now

1.15Pn sin

= 1.1512sin 45

=1.2913

Therefore,

Fmin 1.2913

Problem 11.5

Two helical gears have 20 and 34 teeth and a normal diametral pitch of 8. The left-handed pinion has a helix angle of 40˚ and a rotational speed of 1000 rpm. The gear is also left handed and has a helix angle of 40˚. Determine the angular velocity of the gear, transverse diametral pitch of each gear, and pitch diameters.

Solution

The angular velocity of the gear is given by Eq. (11.11) as

2

3=N3N2

Then,

3 = 2N2N3

=1000 3420

=1700 rpm

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The transverse diametral pitch (Pt) for each gear is related to the normal diametral pitch by Eq. (11.5). Then

Pt = Pncos = 8cos40 = 6.128

The pitch diameter for the pinion is given by

d2 =N2Pt= 206.128

= 3.263 in

And the pitch diameter for the gear is given by

d3 =N3Pt= 346.128

= 5.548 in

Problem 11.6

Two standard spur gears have a diametral pitch of 10, a pressure angle of 20˚, and a velocity ratio equal to 3.5:1. The center distance is 8.55 in. Two helical gears are to be used to replace the two spur gears such that the center distance and angular velocity ratio remain unchanged. The helical gears are also to be cut with the same hob as that used to cut the spur gears. Determine the helix angle, tooth numbers, and minimum face width for the new gears if the helix angle is kept to a minimum.

Solution

For the helical gears, we need to determine the helix angle and the number of teeth on each gear. We will find the number of teeth and transverse diametral pitch first and then determine the helix angle. The velocity ratio is given by Eq. (11.11) as

2

3=rp3rp2

=Dp3Dp2

=N3 / PtN2 / Pt

=N3N2

and the center distance is given by Eq. (11.12)

C = r2 + r3 =D2 +D32

For the spur gear,

d3 =N3N2d2 =3.5d2

and

C = 8.55 = d2 +d32

=d2 +3.5d2

2=4.5d22

= 2.25d2

Therefore,

d2 = 3.8 inand

d3 = 3.5d2 =13.3 inAlso,

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N2 = Pnd2 =10(3.8) = 38and

N3 = Pnd3 =10(13.3) =133

The pitch radii for the helical gears must be the same as the corresponding radii for the spur gears. However, the tooth numbers can and will be different. The transverse diametral pitch is related to the normal diametral pitch by

Pt = Pncos (11.5)

and to the teeth numbers by

Pt =N22rp2

=N32rp3

(11.19)

From Eq. (11.5), it is clear that Pt < Pn . Therefore, based on Eq. (11.19), the tooth numbers on the helical gears must be less than those on the spur gears. As a result, when we investigate tooth numbers that satisfy Eq. (11.18), we need only consider values which are lower than the corresponding values for the spur gears. A set of values is

N2 N3 = 3.5N238 133 (spur gear)37 129.536 126

From the table, the first set of teeth numbers that are integers are N2 = 36 and N3 = 126. For these numbers, the transverse pitch is given by Eq. (11.19) as

Pt =N22rp2

=N32rp3

= 362(1.8)

= 1262(6.65)

= 9.474

From Eq. (11.5),

= cos 1 PtPn

= cos 1 9.474

10( ) =18.672̊

Notice that this is the lowest helix angle possible (other than 0) if the center distance and velocity ratio are to be maintained. The minimum face width is given by Eq. (11.14):

F1.15pttan

or F 1.15Pt tan

Therefore,

F1.15

9.474tan(18.672)=1.128 in( )

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Problem 11.7

Two standard spur gears have a diametral pitch of 16 and a pressure angle of 20˚. The tooth numbers are 36 and 100, and the gears were meshed at a standard center distance. After the gear reducer was designed and tested, the noise of the drive was found to be excessive. Therefore, the decision was made to replace the spur gears with helical gears. The helix angle chosen was 22˚, and the tooth numbers were to remain unchanged. Determine the change in center distance required.

Solution

The center distance for the spur gears is given by Eq. (11.12)

Cc = r2 + r3 =D2 +D32

=N2 + N32Pn

= 36 +1002(16)

= 4.25 in

For the helical gears,

Ch =dt2 + dt32

=N2 + N32Pt

=N2 + N32Pncos

=Cccos

= 4.25cos22

= 4.583 in

The change in center distance is given by

C = Ch Cc = 4.583 4.25 = 0.333 in

Problem 11.8

A spur gear transmission consists of a pinion that drives two gears. The pinion has 24 teeth and a diametral pitch of 12. The velocity ratio for the pinion and one gear is 3:2 and for the pinion and the other gear is 5:2. To reduce the noise level, all three gears are to be replaced by helical gears such that the center distances and velocity ratios remain the same. The helical gears will be cut with a 16 pitch, 20˚ hob. If the helix angle is kept as low as possible, determine the number of teeth, face width, hand, helix angle, and outside diameter for each of the gears.

Solution

Assume that the gear arrangement is as shown in the figure.

2

34

For the helical gears, we need to determine the helix angle and the number of teeth on each gear. We will find the number of teeth and transverse diametral pitch first and then determine the helix angle. For the pinion,

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d2 =N2Pn= 2412

= 2 in

The velocity ratio for the first gears is given by Eq. (11.11) as

2

3=rp3rp2

=dt3dt2

=N3 / PtN2 / Pt

=N3N2

= 32

Therefore,

N3 = 32N2 = 32

24 = 36

and

dt3 =32dt2 =

32(2) = 3 in

For the second gear,

2

4=rp4rp2

=dt4dt2

=N4 / PtN2 / Pt

=N4N2

= 52

Therefore,

N4 = 52N2 = 52

24 = 60

and

dt4 =52dt2 =

52(2) = 5 in

The center distances is given by Eq. (11.12). Then

C1 = r2 + r3 =dt2 + dt32

= 2 +32

= 2.5 in

and

C2 = r2 + r4 =dt2 + dt42

= 2 + 52

= 3.5 in

The pitch radii for the helical gears must be the same as the corresponding radii for the spur gears. However, the tooth numbers can and will be different. The transverse diametral pitch is related to the normal diametral pitch by

Pt = Pncos (11.5)


Pt =N22rp2

=N32rp3

(11.19)

From Eq. (11.5), it is clear that Pt < Pn . Therefore, based on Eq. (11.19), the tooth numbers on the helical gears must be less than those on the spur gears. As a result, when we investigate tooth numbers that satisfy Eq. (11.18), we need only consider values which are lower than the corresponding values for the spur gears. A set of values is

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N2 N4 =1.5N2 N4 = 2.5N224 36 60(spur gear)23 34.5 57.522 33 55

From the table, the first set of teeth numbers that are integers for the three gears are N2 = 22 and N3 = 33, and N4 = 55. For these numbers, the transverse pitch is given by Eq. (11.19) as

Pt =N22rp2

=N32rp3

=N42rp4

= 222(1)

= 332(1.5)

= 552(2.5)

=11

From Eq. (11.5),

= cos 1 PtPn

= cos 1 11

16( ) = 46.567̊

Notice that this is the lowest helix angle possible (other than 0) if the center distance and velocity ratio are to be maintained. The minimum face width is given by Eq. (11.14):

F1.15pttan

or F 1.15Pt tan

Therefore,

F1.15

11tan(46.567)= 0.311 in( )

The blank diameters of the three gears are given by

do2 = dp2 + 2a2 = dp2 + 2kPn= 2 +2 1

16= 2.125 in ,

do3 = dp3 + 2a3 = dp3 + 2kPn= 3+ 2 1

16= 3.125 in

and

do4 = dp4 + 2a4 = dp4 +2kPn= 5 +2 1

16= 5.125 in

The hand of the gears is arbitrary; however, the pinion will be one hand and the two gears will be the opposite hand. For example, if the pinion is right handed, gears 3 and 4 will be left handed.

Problem 11.9

A pair of helical gears have a module in the normal plane of 3 mm, a normal presure angle of 20˚, and a helix angle of 45˚. The gears mesh with parallel shafts and have 30 and 48 teeth. Determine the transverse module, the pitch diameters, the center distance, and the minimum face width.

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Solution


t = tan 1 tan n

cos

= tan 1 tan20̊

cos45̊( ) = tan 1 0.5147( ) = 27.24̊

The transverse module (mt) for both gears is related to the normal module by

mt =dtN=

mncos

= 3cos45

= 4.243mm

The transverse pitch diameters are

dt2 = N2mt = 30(4.243) =127.28mmand

dt3 = N3mt = 48(4.243) = 203.66 mm

The center distance is given by

C = r2 + r3 =dt2 + dt32

= 127.28 +203.662

=165.47mm


F1.15pnsin

= 1.15 mnsin

= 1.15 3sin 45

= 15.328mm

Problem 11.10

Two 20˚ spur gears have 36 and 90 teeth and a module of 1.5. The spur gears are to be replaced by helical gears such that the center distance and velocity ratio are not changed. The maximum allowed face width is 12.7 mm, and the hob module is 1.5 mm. Design the helical gear pair that has the smallest helix angle possible. Determine the numbers of teeth, the face width, the helix angle, and the outside diameters of the gears.

Solution

For the helical gears, we need to determine the helix angle and the number of teeth on each gear. We will find the number of teeth and transverse diametral pitch first and then determine the helix angle. The velocity ratio is given by Eq. (11.11) as

2

3=rp3rp2

=dp3dp2

=N3 / PtN2 / Pt

=N3N2

= 9036

= 2.5 (1)

and the center distance is given by Eq. (11.12)

C = r2 + r3 =d2 + d32

For the spur gear,

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d2 = N2m = 36(1.5) = 54 mm , (2)

d3 = N3m = 90(1.5) = 135mm , (3)

Assume that full depth gears are used. Then the addendum length is given by

a = 1Pn=mn

The outside diameter of the gears are given by

do2 = d2 + 2mn = 54 +2(1.5) = 57mm

and

do3 = d3+ 2mn = 135+ 2(1.5) = 138mm

The center distance is given by

C =d2 +d32

=54 +135

2= 94.5mm

(4)

The pitch radii for the helical gears must be the same as the corresponding radii for the spur gears. However, the tooth numbers can and will be different. The transverse module is related to the normal module by

mt = mn / cos (5)


mt =dt2N2

=dt3N3

(6)

From Eq. (5), it is clear that mt >mn . Therefore, based on Eq. (6), the tooth numbers on the helical gears must be less than those on the spur gears. As a result, when we investigate tooth numbers that satisfy Eq. (1), we need only consider values which are lower than the corresponding values for the spur gears. A set of values is

N2 N3 = 2.5N236 90 (spur gear)35 87.534 8533 82.532 8031 77.530 75

From the table, the first set of teeth numbers that are integers are N2 = 34 and N3 = 85. For these numbers, the transverse module is given by

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mt =dp2N2

=dp3N3

= 5434

=1.588mm

From Eq. (5), the helix angle is given by

= cos 1 mnmt

= cos 1 1.5

1.588( ) =19.19̊

Notice that this is the lowest helix angle possible (other than 0) if the center distance and velocity ratio are to be maintained. The minimum face width is given by

F 1.15 mttan

Therefore,

F1.15 mttan

= 1.15 1.588tan(19.19)

=16.486 mm

This exceeds the maximum allowed face width of 12.7 mm. Therefore, we must try the next lower values for the tooth numbers. These are N2 = 32 and N3 = 80. For these numbers, the transverse module is given by

mt =dp2N2

=dp3N3

= 5432

=1.688mm

From Eq. (5), the helix angle is given by

= cos 1 mnmt

= cos 1 1.5

1.688( ) = 27.266̊

The minimum face width is given by

F1.15 mttan

= 1.15 1.688tan(27.266)

=11.833 mm

Problem 11.11

Two helical gears are cut with a 20˚ hob with a module of 2. One gear is right handed, has a 30˚ helix angle, and has 36 teeth. The second gear is left handed, has a 40˚ helix angle, and has 72 teeth. Determine the shaft angle, the angular velocity ratio, and the center distance.

Solution

The shaft angle is given by Eq. (11.21). Then,

= 2 ± 3 = 30̊ 40̊ = 10˚

The angular velocity of the gear is given by Eq. (11.11) as

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2

3=N3N2

Then,

3

2=N2N3

= 3672

= 2

The pitch diameter for gear 2 is given by

dt2 =N2mncos

=36(2)cos(30)

= 83.14mm

and for gear 3,

dt2 =N2mncos

=72(2)cos(40)

=187.98mm

The center distance is

C = r2 + r3 =d2 + d32

= 83.14 +187.982

=135.56 mm

Problem 11.12

Two crossed shafts are connected by helical gears such that the velocity ratio is 3:1, and the shaft angle is 60˚. The center distance is 10 in, and the normal diametral pitch is 8. The pinion has 35 teeth. Assume that the gears are the same hand and determine the helix angles, pitch diameters, and recommended face widths.

Solution

The velocity ratio is given by

2

3=N3N2

= 3

Therefore,

N3 = 3N2 = 3(35) =105

The pitch diameters for the two gears are given by

dp2 =N2Pt2

=N2

Pncos 2(1)

and

dp3 =N3Pt3

=N3

Pncos 3(2)


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C = r2 + r3 =d2 + d32

= 12Pn

N2cos 2

+ N3cos 3

=10 in

Or,N2

cos 2+

N3cos 3

=10(2)Pn

and35

cos 2+ 105cos 3

= 2(10)(8) =160 (3)

Both gears are of the same hand. Therefore, the shaft angle is given by

= 2 + 3 = 60̊ (4)

To find the helix angles, we must solve Eqs. (3) and (4) simultaneously. First use Eq. (4) to solve for 2 in terms of 3 . Then substitute the result into Eq. (3). This gives

35cos(60 3)

+ 105cos 3

= 160

This equation can be solved iteratively for 3 using MATLAB. The result is

2 = 32.298̊and

3 = 27.702̊

The pitch diameters are given by

dp2 =N2

Pncos 2= 358cos(32.298̊ )

= 5.176 in

and

dp3 =N2

Pncos 3= 1058cos(27.702̊ )

=14.182 in

The minimum face width for each gear is given by

F1.15pnsin

= 1.15Pnsin

Then,

F21.15Pn sin 2

= 1.158sin(32.298̊ )

= 0.845 in[ ]and

F31.15Pnsin 3

= 1.158sin(27.702˚)

= 0.971in[ ]The recommended face widths would be 0.85 in and 1.00 in for gears 2 and 3, respectively.

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Problem 11.13

Two crossed shafts are connected by helical gears such that the velocity ratio is 3:2, and the shaft angle is 90˚. The center distance is 5 in. Select a pair of gears that will satisfy the design constraints. What other information might be considered to reduce the number of arbitrary choices for the design?

Solution

To complete the design, we need to select diametral pitch for the hob used to cut the gears and the number of teeth on one of the gears. We also need to identify if both gears are the same hand or opposite hand.

Initially assume that the helix angles on the gears are the same hand. Also select a hob with a diametral pitch of 12 and assume that the pinion has 30 teeth. If these values do not give helix angles which are approximately equal, we can select other values.


2

3=N3N2

= 32= 1.5

Therefore,

N3 =1.5N2 =1.5(30) = 45


dp2 =N2Pt2

=N2

Pncos 2(1)

and

dp3 =N3Pt3

=N3

Pncos 3(2)


C = r2 + r3 =d2 + d32

= 12Pn

N2cos 2

+ N3cos 3

= 5 in

Or,N2

cos 2+

N3cos 3

= 5(2)Pn

and30

cos 2+ 45cos 3

= 2(5)(12) =120 (3)

Assume that both gears are of the same hand. Therefore, the shaft angle is given by

= 2 + 3 = 90̊ (4)

To find the helix angles, we must solve Eqs. (3) and (4) simultaneously. First use Eq. (4) to solve for 2 in terms of 3 . Then substitute the result into Eq. (3). This gives

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30cos(90 3)

+ 45cos 3

= 120

This equation can be solved iteratively for 3 using MATLAB. The result is

-0.00001 32.149 57.851 2.953 7.047 0.566 0.356

2 = 32.149̊and

3 = 57.851̊


dp2 =N2

Pncos 2= 3012cos(32.149̊ )

= 2.953 in

and

dp3 =N2

Pncos 3= 3012cos(57.851̊ )

= 7.047 in

The minimum face width for each gear is given by

F1.15pnsin

= 1.15Pnsin

Then,

F21.15Pn sin 2

= 1.1512sin(32.149̊ )

= 0.566 in[ ]and

F31.15Pnsin 3

= 1.1512sin(57.851̊ )

= 0.356 in[ ]The recommended face widths would be 0.60 in and 0.40 in for gears 2 and 3, respectively.

To reduce the number of arbitrary choices we could specify limits on the helix angles or on the tooth numbers. Also, we have not considered strength yet. The stress and wear equations for the gear teeth will add additional constraints.

Problem 11.14

A helical gear with a normal diametral pitch of 8 is to be used to drive a spur gear at a shaft angle of 45˚. The helical gear has 21 teeth, and the velocity ratio is 2:1. Determine the helix angle for the helical gear and the pitch diameter of both gears.

Solution

The helix angle for the spur gear is 0; therefore, the helix angle for the helical gear must be the same as the shaft angle or 45˚. Both gears must have the same normal diametral pitch. The helical gear is the pinion (gear 2) and the gear is gear 3.

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2

3=N3N2

= 32= 1.5

Therefore,

N3 = 2N2 = 2(21) = 42


dp2 =N2Pt2

=N2

Pncos 2= 21Pncos45

= 3.712 in (1)

and

dp3 =N3Pt3

=N3

Pncos 3= 418cos0

= 5.125 in (2)

Problem 11.15

Two crossed helical gears connect shafts making an angle of 45˚. The pinion is right handed, has a helix angle of 20˚, and contains 30 teeth. The gear is also right handed and contains 45 teeth. The transverse diametral pitch of the gear is 5. Determine the pitch diameter, the normal pitch, and the lead for each gear.

Solution

Because both gears are of the same hand, the relationship between the helix angles and shaft angle is

= 2 + 3 = 45̊

Therefore,

3 = 45̊ 2 = 45̊ 20 =̊ 35̊

The pitch diameters for the gear is given by

dp3 =N3Pt3

= 455= 9 in

and the normal diametral pitch for both gears is

Pn =Pt3

cos 3= 5cos35̊

= 6.104


pn = Pn=6.104

= 0.515 in

This is a nonstandard pitch and would require a special hob. The pitch diameter for the pinion is

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dp3 =N3Pt3

=N3

Pncos 3


dp2 =N2

Pncos 2= 306.104cos(20̊ )

= 5.230 in

The axial pitches for the two gears are

pa2 =pn

sin 2= 0.515sin(20̊ )

=1.505 in

and

pa3 =pn

sin 3= 0.515sin(35̊ )

= 0.897 in

The leads for the two gears are

L2 = N2pa2 = 30(1.505) = 45.15and

L3 = N3pa3 = 45(0.897) = 40.365

Problem 11.16

The worm of a worm gear set has 2 teeth, and the gear has 58 teeth. The worm axial pitch is 1.25 in, and the pitch diameter is 3 in. The shaft angle is 90˚. Determine the center distance for the two gears, the helix angle, and the lead for the worm.

Solution

Because the shaft angle is 90˚, the transverse circular pitch of the gear is equal to the axial pitch of the worm. Therefore,

pt3 =1.25

Therefore, the pitch diameter of the gear is

dt3 =pt3N3 = 1.25(58) = 23.078 in


C = r2 + r3 =d2 + d32

= 3+ 23.0782

=13.039 in

To determine the helix angles, use the relationships for the shaft angle and angular velocity ratio. Because both gears are of the same hand, the relationship between the helix angles and shaft angle is

= 2 + 3 = 90̊ (1)

The angular velocity ratio is

- 451 -

2

3=N3N2

=dt3 cos 3

dt2 cos 2=dt3 cos 3

dt2 cos 2

Then,

582=23.078cos 3

3cos 2

and

cos 2 = 0.26526cos 3 (2)

Combining Eqs. (1) and (2),

cos(90 3) 0.26526cos 3 = 0 (3)

Equation (3) can be solved iteratively for 3 by using MATLAB. The result is

2 = 75.144˚and

3 =14.856̊

The lead for the worm is given by

L2 = N2pa2 = 2(1.25) = 2.5

Problem 11.17

The shaft angle between two shafts is 90˚, and the shafts are to be connected through a worm gear set. The center distance is 3 in, and the velocity ratio is 30:1. Determine a worm and gear that will satisfy the design requirements. Specify the number of teeth, lead angle, and pitch diameter for each gear. Also, determine the face width for the gear.

Solution

This problem has considerable design latitude. We need to satisfy only three equations. One equation is for the shaft angle

= 2 + 3 = 90̊

The second is for the velocity ratio

2

3=N3N2

=dt3 cos 3

dt2 cos 2=dt3 cos 3

dt2 cos 2

and the third is for the center distance

- 452 -

C = rt2 + rt3 =N22Pt2

+N32Pt3

=N2

2Pncos 2+

N32Pncos 3

= 12Pn

N2cos 2

+N3

cos 3

We can rewrite the equations in terms of the helix angles, tooth numbers, and normal diametral pitch. Doing this and substituting for the known variables,

2 + 3 = 90̊ (1)

N3N2

= 301

(2)

12Pn

N2cos 2

+N3

cos 3

= 3 (3)

This gives us three equations in five unknowns. Therefore, we must select values for two of the variables. Let us choose N2 = 2 and Pn= 12. Then from Eq. (2),

N3 = 30N2 = 30(2) = 60

Equations (1) and (2) can then be combined as

12Pn

N2cos 2

+N3

cos 3

= 12(12)

2cos(90 3)

+ 60cos 3

= 3

or2

cos(90 3)+ 60cos 3

= 3(2)(12) = 72 (4)

Equation (4) can be solved iteratively using MATLAB. The result is

2 = 79.504and

3 =10.496

The pitch diameter of the gear is

dt3 =pt3N3 = 1.25(58) = 23.078 in


C = r2 + r3 =d2 + d32

= 3+ 23.0782

=13.039 in

To determine the helix angles, use the relationships for the shaft angle and angular velocity ratio. Because both gears are of the same hand, the relationship between the helix angles and shaft angle is = 2 + 3 = 90̊ (1)

The angular velocity ratio is

- 453 -

2

3=N3N2

=dt3 cos 3

dt2 cos 2=dt3 cos 3

dt2 cos 2

Then,582=23.078cos 3

3cos 2and

cos 2 = 0.26526cos 3 (2)

Combining Eqs. (1) and (2),

cos(90 3) 0.26526cos 3 = 0 (3)

Equation (3) can be solved iteratively for 3 by using MATLAB. The result is

2 = 79.504and

3 =10.496

The transverse pitch diameters are

dt2 =N2

2Pncos 2= 22(12)cos(79.504)

= 0.457 in

and

dt3 =N3

2Pncos 3= 602(12)cos(10.496)

= 2.543 in

The lead for the worm and gear are

L2 = N2pa2 =N2pnsin 2

=N2

Pn sin 2= 212sin(79.504)

= 0.532

and


=N3

Pn sin 3= 212sin(10.496)

= 2.874

The minimum facewidth for the gear is given by

F1.15pnsin

= 1.15Pnsin

Then,

F31.15Pnsin 3

= 1.1512sin(10.496̊ )

= 1.653 in[ ]For this type of gear, the minimum facewidth is not very meaningful. The worm contacts the gear at one point only, so the entire face is not contacted.

- 454 -

Problem 11.18

A worm with two teeth drives a gear with 50 teeth. The gear has a pitch diameter of 8 in and a helix angle of 20˚. The shaft angle between the two shafts is 80˚. Determine the lead and pitch diameter of the worm.

Solution

The normal diametral pitch for the gears is

Pn =N3

2dt3 cos 3= 502(8)cos20̊

= 3.326

The shaft angles for the two gears are related by

= 2 + 3 = 80̊

Therefore,

2 = 80̊ 3 = 80̊ 20 =̊ 60̊

The pitch diameter of the worm is then given by

dt2 =N2

2Pncos 2= 22(3.326)cos(60)

= 0.601 in

The lead for the worm is


=N2

Pn sin 2= 23.326sin(60)

= 2.181

Problem 11.19

Two straight-toothed bevel gears mesh with a shaft angle of 90˚ and a diametral pitch of 5. The pinion has 20 teeth, and the gear ratio is 2:1. The addendum and dedendum are the same as for 20˚ stub teeth. For the gear, determine the pitch radius, cone angle, outside diameter, cone distance, and face width.

Solution

The number of teeth on the gear is given by

N3 = N2 2

3= 202

1= 40

The pitch cone angle for the gear is given by Eq. (11.27). Then,

3 = tan 1 N3N2[ ]= tan 1(2) = 63.435̊

- 455 -

The pitch cone angle for the pinion is given Eq. (11.31). Then,

2 = 3 = 90̊ 63.435̊ = 26.565̊

The pitch radius for the large end of the gear is given by

r3 =N32Pn

= 402(5)

= 4

The cone distance is given by

ro =d2

2sin 2=

d32sin 3

=r3

sin 3= 4sin(63.435)

= 4.472 in

The outside diameter at the back of the gear is

do3 = d3 +2a = d3 + 2kPn= 4 +2 0.8

5= 4.32 in

The face width limit is given by Eq. (11.43). Then

F < 0.3ro = 0.3(4.472) =1.342[ ]

or

F < 10Pn= 105= 2[ ]

We should use the smaller value for F or

F = 2 in

Problem 11.20

A pair of straight-toothed bevel gears mesh with a shaft angle of 90˚ and a diametral pitch of 6. The pinion has 18 teeth, and the gear ratio is 2:1. The addendum and dedendum are the same as for 20˚ full-depth spur-gear teeth. Determine the number of teeth on the gear and the pitch diameters of both the pinion and gear. Also, for the gear, determine the pitch-cone angle, outside diameter, cone distance, and face width.

Solution


N3 = N2 2

3=182

1= 36

The pitch cone angle for the gear is given by Eq. (11.27). Then,

3 = tan 1 N3N2[ ]= tan 1(2) = 63.435̊

- 456 -

The pitch-cone angle for the pinion is given Eq. (11.31). Then,

2 = 3 = 90̊ 63.435̊ = 26.565̊

The pitch diameter for the large end of the gear is given by

d3 =N3Pn= 36(6)

= 6 in

The pitch diameter for the pinion is given by Eq. (11.32). Then

d2 = d3N2N3

= 61836

= 3in

The cone distance is given by

ro =d2

2sin 2=

d32sin 3

=r3

sin 3= 3sin(63.435)

= 3.354 in

The outside diameter at the back of the gear is

do3 = d3 +2a = d3 + 2kPn= 6 +2 1

6= 6.333 in

The face width limit is given by Eq. (11.43). Then

F < 0.3ro = 0.3(3.354) =1.006[ ]and

F < 10Pn= 106=1.666[ ]

We should use the smaller value for F or

F =1in

Problem 11.21

A pair of straight-toothed bevel gears mesh with a shaft angle of 80˚ and a diametral pitch of 7. The pinion has 20 teeth and a pitch cone angle of 40˚. The gear ratio is 3:2. Determine the number of teeth on the gear and the pitch diameters of both the pinion and gear. Also, determine the equivalent spur gear radii for both the pinion and the gear.

Solution


N3 = N2 2

3= 20 3

2= 30

The pitch cone angle for the gear is given by Eq. (11.31).

- 457 -

3 = 2 = 80̊ 40̊ = 40̊

The pitch diameter for the large end of the pinion is given by

d2 =N2Pn= 207= 2.857 in

and for the gear,

d3 =N3Pn= 307= 4.286 in

The equivalent spur gear radius for the pinion is given by Eq. (11.39) as

re2 =d2

2cos 2= 2.8572cos(40)

=1.865 in

and for the gear,

re3 =d3

2cos 3= 4.2862cos(40)

= 2.797 in

Problem 11.22

A pair of straight-toothed bevel gears mesh with a shaft angle of 45˚ and a module of 5.08. The pinion has 16 teeth and a pitch cone angle of 20˚. The gear ratio is 3:2. Determine the number of teeth on the gear and the pitch diameters of both the pinion and gear. Also determine the back-cone distance and the back-cone angle for the gear.

Solution


N3 = N2 2

3=16 3

2= 24

The pitch cone angle for the gear is given by Eq. (11.31).

3 = 2 = 45̊ 20̊ = 25̊

The pitch diameter for the large end of the pinion is given by

d2 = N2mn =16(5.08) = 81.28mm

and for the gear,

d3 = N3mn = 24(5.08) =121.92mm

The back-cone radius for the gear is given by Eq. (11.39) as

- 458 -

re3 =d3

2cos 3= 121.922cos(40)

= 79.58mm

The back-cone angle is given by

= 2sin 1 r3re3( ) = 2sin 1 121.92

2(79.58)[ ] =100̊

solutions to chapter 11 exercise problems problem...

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